Creating a Star to Heart Animation with SVG and Vanilla JavaScript

In my previous article, I’ve shown how to smoothly transition from one state to another using vanilla JavaScript. Make sure you check that one out first because I’ll be referencing some things I explained there in a lot of detail, like demos given as examples, formulas for various timing functions or how not to reverse the timing function when going back from the final state of a transition to the initial one.

The last example showcased making the shape of a mouth to go from sad to glad by changing the d attribute of the path we used to draw this mouth.

Manipulating the path data can be taken to the next level to give us more interesting results, like a star morphing into a heart.

The star to heart animation we’ll be coding.

The idea

Both are made out of five cubic Bézier curves. The interactive demo below shows the individual curves and the points where these curves are connected. Clicking any curve or point highlights it, as well as its corresponding curve/point from the other shape.

Note that all of these curves are created as cubic ones, even if, for some of them, the two control points coincide.

The shapes for both the star and the heart are pretty simplistic and unrealistic ones, but they’ll do.

The starting code

As seen in the face animation example, I often choose to generate such shapes with Pug, but here, since this path data we generate will also need to be manipulated with JavaScript for the transition, going all JavaScript, including computing the coordinates and putting them into the d attribute seems like the best option.

This means we don’t need to write much in terms of markup:

<svg>
<path id='shape'/>
</svg>

In terms of JavaScript, we start by getting the SVG element and the path element – this is the shape that morphs from a star into a heart and back. We also set a viewBox attribute on the SVG element such that its dimensions along the two axes are equal and the (0,0) point is dead in the middle. This means the coordinates of the top left corner are (-.5*D,-.5*D), where D is the value for the viewBox dimensions. And last, but not least, we create an object to store info about the initial and final states of the transition and about how to go from the interpolated values to the actual attribute values we need to set on our SVG shape.

Now that we got this out of the way, we can move on to the more interesting part!

The geometry of the shapes

The initial coordinates of the end points and control points are those for which we get the star and the final ones are the ones for which we get the heart. The range for each coordinate is the difference between its final value and its initial one. Here, we also rotate the shape as we morph it because we want the star to point up and we change the fill to go from the golden star to the crimson heart.

Alright, but how do we get the coordinates of the end and control points in the two cases?

Star

In the case of the star, we start with a regular pentagram. The end points of our curves are at the intersection between the pentagram edges and we use the pentagram vertices as control points.

Regular pentagram with vertex and edge crossing points highlighted as control points and end points of five cubic Bézier curves (live).

Getting the vertices of our regular pentagram is pretty straightforward given the radius (or diameter) of its circumcircle, which we take to be a fraction of the viewBox size of our SVG (considered here square for simplicity, we’re not going for tight packing in this case). But how do we get their intersections?

First of all, let’s consider the small pentagon highlighted inside the pentagram in the illustration below. Since the pentagram is regular, the small pentagon whose vertices coincide with the edge intersections of the pentagram is also regular. It also has the same incircle as the pentagram and, therefore, the same inradius.

So if we compute the pentagram inradius, then we also have the inradius of the inner pentagon, which, together with the central angle corresponding to an edge of a regular pentagon, allows us to get the circumradius of this pentagon, which in turn allows us to compute its vertex coordinates and these are exactly the edge intersections of the pentagram and the endpoints of our cubic Bézier curves.

Our regular pentagram is represented by the Schläfli symbol{5/2}, meaning that it has 5 vertices, and, given these 5 vertex points equally distributed on its circumcircle, 360°/5 = 72° apart, we start from the first, skip the next point on the circle and connect to the second one (this is the meaning of the 2 in the symbol; 1 would describe a pentagon as we don’t skip any points, we connect to the first). And so on – we keep skipping the point right after.

In the interactive demo below, select either pentagon or pentagram to see how they get constructed.

This way, we get that the central angle corresponding to an edge of the regular pentagram is twice of that corresponding to the regular pentagon with the same vertices. We have 1·(360°/5) = 1·72° = 72° (or 1·(2·π/5) in radians) for the pentagon versus 2·(360°/5) = 2·72° = 144° (2·(2·π/5) in radians) for the pentagram. In general, given a regular polygon (whether it’s a convex or a star polygon doesn’t matter) with the Schläfli symbol {p,q}, the central angle corresponding to one of its edges is q·(360°/p) (q·(2·π/p) in radians).

We also know the pentagram circumradius, which we said we take as a fraction of the square viewBox size. This means we can get the pentagram inradius (which is equal to that of the small pentagon) from a right triangle where we know the hypotenuse (it’s the pentagram circumradius) and an acute angle (half the central angle corresponding to the pentagram edge).

Computing the inradius of a regular pentagram from a right triangle where the hypotenuse is the pentagram circumradius and the acute angle between the two is half the central angle corresponding to a pentagram edge (live).

The cosine of half the central angle is the inradius over the circumradius, which gives us that the inradius is the circumradius multiplied with this cosine value.

Now that we have the inradius of the small regular pentagon inside our pentagram, we can compute its circumradius from a similar right triangle having the circumradius as hypotenuse, half the central angle as one of the acute angles and the inradius as the cathetus adjacent to this acute angle.

The illustration below highlights a right triangle formed from a circumradius of a regular pentagon, its inradius and half an edge. From this triangle, we can compute the circumradius if we know the inradius and the central angle corresponding to a pentagon edge as the acute angle between these two radii is half this central angle.

Computing the circumradius of a regular pentagon from a right triangle where it’s the hypotenuse, while the catheti are the inradius and half the pentagon edge and the acute angle between the two radii is half the central angle corresponding to a pentagon edge (live).

Remember that, in this case, the central angle is not the same as for the pentagram, it’s half of it (360°/5 = 72°).

Good, now that we have this radius, we can get all the coordinates we want. They’re the coordinates of points distributed at equal angles on two circles. We have 5 points on the outer circle (the circumcircle of our pentagram) and 5 on the inner one (the circumcircle of the small pentagon). That’s 10 points in total, with angles of 360°/10 = 36° in between the radial lines they’re on.

The end and control points are distributed on the circumradius of the inner pentagon and on that of the pentagram respectively (live).

We know the radii of both these circles. The radius of the outer one is the regular pentagram circumradius, which we take to be some arbitrary fraction of the viewBox dimension (.5 or .25 or .32 or whatever value we feel would work best). The radius of the inner one is the circumradius of the small regular pentagon formed inside the pentagram, which we can compute as a function of the central angle corresponding to one of its edges and its inradius, which is equal to that of the pentagram and therefore we can compute from the pentagram circumradius and the central angle corresponding to a pentagram edge.

So, at this point, we can generate the path data that draws our star, it doesn’t depend on anything that’s still unknown.

So let’s do that and put all of the above into code!

We start by creating a getStarPoints(f) function which depends on an arbitrary factor (f) that’s going to help us get the pentagram circumradius from the viewBox size. This function returns an array of coordinates we later use for interpolation.

Within this function, we first compute the constant stuff that won’t change as we progress through it – the pentagram circumradius (radius of the outer circle), the central (base) angles corresponding to one edge of a regular pentagram and polygon, the inradius shared by the pentagram and the inner pentagon whose vertices are the points where the pentagram edges cross each other, the circumradius of this inner pentagon and, finally, the total number of distinct points whose coordinates we need to compute and the base angle for this distribution.

After that, within a loop, we compute the coordinates of the points we want and we push them into the array of coordinates.

To compute the coordinates of our points, we use the radius of the circle they’re on and the angle of the radial line connecting them to the origin with respect to the horizontal axis, as illustrated by the interactive demo below (drag the point to see how its Cartesian coordinates change):

In our case, the current radius is the radius of the outer circle (pentagram circumradius RCO) for even index points (0, 2, …) and the radius of the inner circle (inner pentagon circumradius RCI) for odd index points (1, 3, …), while the angle of the radial line connecting the current point to the origin is the point index (i) multiplied with the base angle for point distribution (BAD, which happens to be 36° or π/10 in our particular case).

Since we’ve chosen a pretty big value for the viewBox size, we can safely round the coordinate values so that our code looks cleaner, without decimals.

As for pushing these coordinates into the points array, we do this twice when we’re on the outer circle (the even indices case) because that’s where we actually have two control points overlapping, but only for the star, so we’ll need to move each of these overlapping points into different positions to get the heart.

Next, we put data into our object O. For the path data (d) attribute, we store the array of points we get when calling the above function as the initial value. We also create a function for generating the actual attribute value (the path data string in this case – inserting commands in between the pairs of coordinates, so that the browser knows what to do with those coordinates). Finally, we take every attribute we have stored data for and we set its value to the value returned by the previously mentioned function:

This is a promising start. However, we want the first tip of the generating pentagram to point down and the first tip of the resulting star to point up. Currently, they’re both pointing right. This is because we start from 0° (3 o’clock). So in order to start from 6 o’clock, we add 90° (π/2 in radians) to every current angle in the getStarPoints() function.

ca = i*BAD + .5*Math.PI

This makes the first tip of the generating pentagram and resulting star to point down. To rotate the star, we need to set its transform attribute to a half circle rotation. In order to do so, we first set an initial rotation angle to -180. Afterwards, we set the function that generates the actual attribute value to a function that generates a string from a function name and an argument:

Heart

Since we have the star, let’s next see how we can get the heart!

We start with two intersecting circles of equal radii, both a fraction (let’s say .25 for the time being) of the viewBox size. These circles intersect in such a way that the segment connecting their central points is on the x axis and the segment connecting their intersection points is on the y axis. We also take these two segments to be equal.

We start with two circles of equal radius whose central points are on the horizontal axis and which intersect on the vertical axis (live).

Next, we draw diameters through the upper intersection point and then tangents through the opposite points of these diameters. These tangents intersect on the y axis.

Constructing diameters through the upper intersection point and tangents to the circle at the opposite ends of these diameters, tangents which intersect on the vertical axis (live).

The upper intersection point and the diametrically opposite points make up three of the five end points we need. The other two end points split the outer half circle arcs into two equal parts, thus giving us four quarter circle arcs.

Highlighting the end points of the cubic Bézier curves that make up the heart and the coinciding control points of the bottom one of these curves (live).

Both control points for the curve at the bottom coincide with the intersection of the the two tangents drawn previously. But what about the other four curves? How can we go from circular arcs to cubic Bézier curves?

We don’t have a cubic Bézier curve equivalent for a quarter circle arc, but we can find a very good approximation, as explained in this article.

The gist of it is that we start from a quarter circle arc of radius R and draw tangents to the end points of this arc (N and Q). These tangents intersect at P. The quadrilateral ONPQ has all angles equal to 90° (or π/2), three of them by construction (O corresponds to a 90° arc and the tangent to a point of that circle is always perpendicular onto the radial line to the same point) and the final one by computation (the sum of angles in a quadrilateral is always 360° and the other three angles add up to 270°). This makes ONPQ a rectangle. But ONPQ also has two consecutive edges equal (OQ and ON are both radial lines, equal to R in length), which makes it a square of edge R. So the lengths of NP and QP are also equal to R.

The control points of the cubic curve approximating our arc are on the tangent lines NP and QP, at C·R away from the end points, where C is the constant the previously linked article computes to be .551915.

Given all of this, we can now start computing the coordinates of the end points and control points of the cubic curves making up our star.

Due to the way we’ve chosen to construct this heart, TO0SO1 (see figure below) is a square since it has all edges equal (all are radii of one of our two equal circles) and its diagonals are equal by construction (we said the distance between the central points equals that between the intersection points). Here, O is the intersection of the diagonals and OT is half the ST diagonal. T and S are on the y axis, so their x coordinate is 0. Their y coordinate in absolute value equals the OT segment, which is half the diagonal (as is the OS segment).

We can split any square of edge length l into two equal right isosceles triangles where the catheti coincide with the square edges and the hypotenuse coincides with a diagonal.

Any square can be split into two congruent right isosceles triangles (live).

Using one of these right triangles, we can compute the hypotenuse (and therefore the square diagonal) using Pythagora’s theorem: d² = l² + l². This gives us the square diagonal as a function of the edge d = √(2∙l) = l∙√2 (conversely, the edge as a function of the diagonal is l = d/√2). It also means that half the diagonal is d/2 = (l∙√2)/2 = l/√2.

Applying this to our TO0SO1 square of edge length R, we get that the y coordinate of T (which, in absolute value, equals half this square’s diagonal) is -R/√2 and the y coordinate of S is R/√2.

In the illustration above, the TBk segments are diameter segments, meaning that the TBk arcs are half circle, or 180° arcs and we’ve split them into two equal halves with the Ak points, getting two equal 90° arcs – TAk and AkBk, which correspond to two equal 90° angles, ∠TOkAk and ∠AkOkBk.

Given that ∠TOkS are 90° angles and ∠TOkAk are also 90° angles by construction, it results that the SAk segments are also diameter segments. This gives us that in the TAkBkS quadrilaterals, the diagonals TBk and SAk are perpendicular, equal and cross each other in the middle (TOk, OkBk, SOk and OkAk are all equal to the initial circle radius R). This means the TAkBkS quadrilaterals are squares whose diagonals are 2∙R.

From here we can get that the edge length of the TAkBkS quadrilaterals is 2∙R/√2 = R∙√2. Since all angles of a square are 90° ones and the TS edge coincides with the vertical axis, this means the TAk and SBk edges are horizontal, parallel to the x axis and their length gives us the x coordinates of the Ak and Bk points: ±R∙√2.

Since TAk and SBk are horizontal segments, the y coordinates of the Ak and Bk points equal those of the T (-R/√2) and S (R/√2) points respectively.

Another thing we get from here is that, since TAkBkS are squares, AkBk are parallel with TS, which is on the y (vertical) axis, therefore the AkBk segments are vertical. Additionally, since the x axis is parallel to the TAk and SBk segments and it cuts the TS, it results that it also cuts the AkBk segments in half.

The TB0CB1 quadrilateral has all angles equal to 90° (∠T since TO0SO1 is a square, ∠Bk by construction since the BkC segments are tangent to the circle at Bk and therefore perpendicular onto the radial lines OkBk at that point; and finally, ∠C can only be 90° since the sum of angles in a quadrilateral is 360° and the other three angles add up to 270°), which makes it a rectangle. It also has two consecutive edges equal – TB0 and TB1 are both diameters of the initial squares and therefore both equal to 2∙R. All of this makes it a square of edge 2∙R.

From here, we can get its diagonal TC – it’s 2∙R∙√2. Since C is on the y axis, its x coordinate is 0. Its y coordinate is the length of the OC segment. The OC segment is the TC segment minus the OT segment: 2∙R∙√2 - R/√2 = 4∙R/√2 - R/√2 = 3∙R/√2.

In the TOkAkDk quadrilaterals, we have that all angles are 90° (right) angles, three of them by construction (∠DkTOk and ∠DkAkOk are the angles between the radial and tangent lines at T and Ak respectively, while ∠TOkAk are the angles corresponding to the quarter circle arcs TAk) and the fourth by computation (the sum of angles in a quadrilateral is 360° and the other three add up to 270°). This makes TOkAkDk rectangles. Since they have two consecutive edges equal (OkT and OkAk are radial segments of length R), they are also squares.

This means the diagonals TAk and OkDk are R∙√2. We already know that TAk are horizontal and, since the diagonals of a square are perpendicular, it results the OkDk segments are vertical. This means the Ok and Dk points have the same x coordinate, which we’ve already computed for Ok to be ±R/√2. Since we know the length of OkDk, we can also get the y coordinates – they’re the diagonal length (R∙√2) with minus in front.

Similarly, in the AkOkBkEk quadrilaterals, we have that all angles are 90° (right) angles, three of them by construction (∠EkAkOk and ∠EkBkOk are the angles between the radial and tangent lines at Ak and Bk respectively, while ∠AkOkBk are the angles corresponding to the quarter circle arcs AkBk) and the fourth by computation (the sum of angles in a quadrilateral is 360° and the other three add up to 270°). This makes AkOkBkEk rectangles. Since they have two consecutive edges equal (OkAk and OkBk are radial segments of length R), they are also squares.

From here, we get the diagonals AkBk and OkEk are R∙√2. We know the AkBk segments are vertical and split into half by the horizontal axis, which means the OkEk segments are on this axis and the y coordinates of the Ek points are 0. Since the x coordinates of the Ok points are ±R/√2 and the OkEk segments are R∙√2, we can compute those of the Ek points as well – they’re ±3∙R/√2.

The coordinates of the newly computed vertices of the TOₖAₖDₖ and AₖOₖBₖEₖ squares (live).

Alright, but these intersection points for the tangents are not the control points we need to get the circular arc approximations. The control points we want are on the TDk, AkDk, AkEk and BkEk segments at about 55% (this value is given by the constant C computed in the previously mentioned article) away from the curve end points (T, Ak, Bk). This means the segments from the endpoints to the control points are C∙R.

In this situation, the coordinates of our control points are 1 - C of those of the end points (T, Ak and Bk) plus C of those of the points where the tangents at the end points intersect (Dk and Ek).

So let’s put all of this into JavaScript code!

Just like in the star case, we start with a getStarPoints(f) function which depends on an arbitrary factor (f) that’s going to help us get the radius of the helper circles from the viewBox size. This function also returns an array of coordinates we later use for interpolation.

Inside, we compute the stuff that doesn’t change throughout the function. First off, the radius of the helper circles. From that, the half diagonal of the small squares whose edge equals this helper circle radius, half diagonal which is also the circumradius of these squares. Afterwards, the coordinates of the end points of our cubic curves (the T, Ak, Bk points), in absolute value for the ones along the horizontal axis. Then we move on to the coordinates of the points where the tangents through the end points intersect (the C, Dk, Ek points). These either coincide with the control points (C) or can help us get the control points (this is the case for Dk and Ek).

Next, we need to put the relevant coordinates into an array and return this array. In the case of the star, we started with the bottom curve and then went clockwise, so we do the same here. For every curve, we push two sets of coordinates for the control points and then one set for the point where the current curve ends.

Note that in the case of the first (bottom) curve, the two control points coincide, so we push the same pair of coordinates twice. The code doesn’t look anywhere near as nice as in the case of the star, but it will have to suffice:

We can now take our star demo and use the getHeartPoints() function for the final state, no rotation and a crimson fill instead. Then, we set the current state to the final shape, just so that we can see the heart:

Ensuring consistent shape alignment

The easiest way to solve this issue is to shift the heart up by an amount depending on the radius of the helper circles:

return [ /* same coords */ ].map(([x, y]) => [x, y - .09*R])

We now have much better alignment, regardless of how we tweak the f factor in either case. This is the factor that determines the pentagram circumradius relative to the viewBox size in the star case (when the default is .5) and the radius of the helper circles relative to the same viewBox size in the heart case (when the default is .25).

Switching between the two shapes

We want to go from one shape to the other on click. In order to do this, we set a direction dir variable which is 1 when we go from star to heart and -1 when we go from heart to star. Initially, it’s -1, as if we’ve just switched from heart to star.

Then we add a 'click' event listener on the _SHAPE element and code what happens in this situation – we change the sign of the direction (dir) variable and we change the shape’s attributes so that we go from a golden star to a crimson heart or the other way around:

Morphing from one shape to another

What we really want however is not an abrupt change from one shape to another, but a gradual one. So we use the interpolation techniques explained in the previous article to achieve this.

We first decide on a total number of frames for our transition (NF) and choose the kind of timing functions we want to use – an ease-in-out type of function for transitioning the path shape from star to heart, a bounce-ini-fin type of function for the rotation angle and an ease-out one for the fill. We only include these, though we could later add others in case we change our mind and want to explore other options as well.

We then specify which of these timing functions we use for each property we transition:

(function init() {
/* same as before */
O.d = {
/* same as before */
tfn: 'ease-in-out'
};
O.transform = {
/* same as before */
tfn: 'bounce-ini-fin'
};
O.fill = {
/* same as before */
tfn: 'ease-out'
};
/* same as before */
})();

We move on to adding request ID (rID) and current frame (cf) variables, an update() function we first call on click, then on every refresh of the display until the transition finishes and we call a stopAni() function to exit this animation loop. Within the update() function, we… well, update the current frame cf, compute a progress k and decide whether we’ve reached the end of the transition and we need to exit the animation loop or we carry on.

We also add a multiplier m variable which we use so that we don’t reverse the timing functions when we go from the final state (heart) back to the initial one (star).

Within the update() function, we want to set the attributes we transition to some intermediate values (depending on the progress k). As seen in the previous article, it’s good to have the ranges between the final and initial values precomputed at the beginning, before even setting the listener, so that’s our next step: creating a function that computes the range between numbers, whether as such or in arrays, no matter how deep and then using this function to set the ranges for the properties we want to transition.

Now all that’s left to do is the interpolation part in the update() function. Using a loop, we go through all the attributes we want to smoothly change from one end state to the other. Within this loop, we set their current value to the one we get as the result of an interpolation function which depends on the initial value(s), range(s) of the current attribute (ini and rng), on the timing function we use (tfn) and on the progress (k):

It’s almost what we wanted – there’s still one tiny issue. For cyclic values like angle values, we don’t want to go back by half a circle on the second click. Instead, we want to continue going in the same direction for another half circle. Adding this half circle from after the second click with the one traveled after the first click, we get a full circle so we’re right back where we started.

We put this into code by adding an optional continuity property and tweaking the updating and interpolating functions a bit: